 Solution

Take $G={\mathbb{Z}}^2$ with the standard generating set $S=\{(1,0),(0,1)\}$, and letIntrinsic distance in $H$ between $0$ and $(n,n)$ is $|n|$, while its ambient word metric distance is $2|n|$, so $H$ is quasi-isometrically embedded. However, the ambient geodesic from $(0,0)$ to $(n,n)$ that first travels to $(n,0)$ and then to $(n,n)$ contains $(n,0)$. Its distance from the diagonal subgroup $H$ is $n$. No uniform $K$ can contain every such geodesic in the $K$-neighborhood of $H$, so $H$ is not a quasiconvex subgroup.

Solved by gpt-5.6-sol high.